Math, asked by chinmay2638, 4 months ago

Construct a triangle ABC with base BC = 7cm, ∠B = 750 and AC+AB = 13 cm​

Answers

Answered by ARRUYADAV
1

Answer:

not possible

Step-by-step explanation:

because angle b is given 750 wich is not possible

Answered by nisha02345
18

Answer:

Given,

Given,to construct a ΔABC where

Given,to construct a ΔABC where BC=7cm,∠B=75 °

& AB+AC=13cm.

& AB+AC=13cm.construction steps :-

& AB+AC=13cm.construction steps :-(i) Draw a line segment BC = 7 cm as a base.

& AB+AC=13cm.construction steps :-(i) Draw a line segment BC = 7 cm as a base.(ii) Now mark ∠B=75 °

and extend it till x.

and extend it till x.(iii) Now, by taking radius 13 cm draw an arc from 'B' on the line of angle just projected.

and extend it till x.(iii) Now, by taking radius 13 cm draw an arc from 'B' on the line of angle just projected.(iv) Now mark it as 'D', & join C to D.

and extend it till x.(iii) Now, by taking radius 13 cm draw an arc from 'B' on the line of angle just projected.(iv) Now mark it as 'D', & join C to D.(v) Draw ⊥

bisectors for DC and mark them as P & Q.

bisectors for DC and mark them as P & Q.(vi) Now extend a line from 'B' to the ⊥

bisectors for DC and mark them as P & Q.(vi) Now extend a line from 'B' to the ⊥ lar

bisectors for DC and mark them as P & Q.(vi) Now extend a line from 'B' to the ⊥ lar bisector that cuts on C & D line.

bisectors for DC and mark them as P & Q.(vi) Now extend a line from 'B' to the ⊥ lar bisector that cuts on C & D line.(vii) Now mark that point as 'A'.

bisectors for DC and mark them as P & Q.(vi) Now extend a line from 'B' to the ⊥ lar bisector that cuts on C & D line.(vii) Now mark that point as 'A'.(viii) Thus, required ΔABC has been formed.

bisectors for DC and mark them as P & Q.(vi) Now extend a line from 'B' to the ⊥ lar bisector that cuts on C & D line.(vii) Now mark that point as 'A'.(viii) Thus, required ΔABC has been formed.(ix) Now, try taking exactly 1/2 of the angle and mark an arc on B.

bisectors for DC and mark them as P & Q.(vi) Now extend a line from 'B' to the ⊥ lar bisector that cuts on C & D line.(vii) Now mark that point as 'A'.(viii) Thus, required ΔABC has been formed.(ix) Now, try taking exactly 1/2 of the angle and mark an arc on B.Conclusion

bisectors for DC and mark them as P & Q.(vi) Now extend a line from 'B' to the ⊥ lar bisector that cuts on C & D line.(vii) Now mark that point as 'A'.(viii) Thus, required ΔABC has been formed.(ix) Now, try taking exactly 1/2 of the angle and mark an arc on B.Conclusion ∴ The required ΔABC will BC = 7 cm,

bisectors for DC and mark them as P & Q.(vi) Now extend a line from 'B' to the ⊥ lar bisector that cuts on C & D line.(vii) Now mark that point as 'A'.(viii) Thus, required ΔABC has been formed.(ix) Now, try taking exactly 1/2 of the angle and mark an arc on B.Conclusion ∴ The required ΔABC will BC = 7 cm,∠B=45

bisectors for DC and mark them as P & Q.(vi) Now extend a line from 'B' to the ⊥ lar bisector that cuts on C & D line.(vii) Now mark that point as 'A'.(viii) Thus, required ΔABC has been formed.(ix) Now, try taking exactly 1/2 of the angle and mark an arc on B.Conclusion ∴ The required ΔABC will BC = 7 cm,∠B=45 & AB + AC = 13 cm is constructed.

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